In part of the proof of the splitting lemma (a left-split short exact sequence of abelian groups is right-split) it seems necessary to invoke the axiom of choice. That is, if $0\to A\overset{f}{\to} B\overset{g}{\to} C\to 0$ is exact and there is a retraction $B\overset{r}{\to}A$, then we can find a section $C\overset{s}{\to}B$ by choosing any right inverse of $g$ and removing the part in the kernel of $f$, which gives a well-defined morphism independent of the choice.

Is this invocation of the axiom of choice essential? I thought I had an example that showed it was: $0\to\mathbb{Q}\to\mathbb{R}\to\mathbb{R}/\mathbb{Q}\to0$ splits on the right if you can choose a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$. But actually now I think it doesn't split on the left without a basis either.

The map $C\to B$ is supposed to be a canonical injection map. Can something be "canonical" if it requires choice?

The sequence splits if and only if $\mathbb R\cong\mathbb R\oplus\mathbb Q$. I know this follows from a Hamel basis of $\mathbb R$ over $\mathbb Q$ but I'm not sure it implies such basis exists.
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Asaf KaragilaMay 30 '12 at 6:05

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I deleted an earlier answer because I missed the second part of the question. The first part can be done without choice: Consider the morphism $h = 1_B-fr\colon B \to B$. Then $hf = f -frf = 0$, so $h$ factors as $h = sg$ for a unique morphism $s\colon C \to B$ (note that $s(b+A) = h(b)$ is well-defined as a map $s: B/A \to B$). Now $gsg = gh =g-gfr = g = 1_C g$, so $gs = 1_C$ because $g$ is onto, so $s$ is a section of $g$.
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t.b.May 30 '12 at 6:10

@t.b.: It might just be the hangover, but either I answered the "hard" part, or I missed something. Have I missed something?
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Asaf KaragilaMay 30 '12 at 6:29

@Asaf: I don't understand why the map $\mathbb{Q} \to \mathbb{R} \oplus \mathbb{Q}$ given by composing $\mathbb{Q} \to \mathbb{R} \to \mathbb{R} \oplus \mathbb{Q}$ should have a left inverse. There is some map having a left inverse, namely the inclusion into the second summand, but why is this the same as the map obtained from composing the natural inclusion with the isomorphism $\mathbb{R} \to \mathbb{R} \oplus \mathbb{Q}$ you assume to exist?
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t.b.May 30 '12 at 6:37

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To be clear, the splitting lemma does not require choice. Or even the law of excluded middle. It's completely constructive.
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Zhen LinMay 30 '12 at 6:52

We also have $frsg = frh = fr-frfr = 0$, so $frs = 0$ because $g$ is onto, so $rs = 0$ because $f$ is injective and by construction of $s$ we have $fr + sg = fr + h = 1_B$. In particular, we have an isomorphism
$$
\begin{bmatrix} f & s \end{bmatrix}\colon A \oplus C \longleftrightarrow {B :} \begin{bmatrix} r \\ g \end{bmatrix}.
$$
Note that $s$ is uniquely determined by $r$ and similarly one shows that if $s$ is a right inverse of $g$ then there is a uniquely determined right inverse $r$ of $g$ such that $fr = 1-sg$.

Concerning the existence of a retraction of the inclusion $\mathbb{Q} \to \mathbb{R}$: assuming such a retraction exists, the previous part gives us a section $s\colon\mathbb{R/Q} \to \mathbb{R}$ of $g\colon \mathbb{R} \to \mathbb{R/Q}$. Modifiying this section by setting $t(x) = s(x) - \lfloor s(x) \rfloor$, we get a Vitali set $t(\mathbb{R}/\mathbb{Q}) \subset [0,1]$, whose existence we cannot prove from ZF alone, so we cannot prove from ZF that there is a left inverse of the inclusion $\mathbb{Q} \to \mathbb{R}$.

A simple example of something “canonical” that requires choice in order to be non-trivial would be a product of an arbitrary collection of sets.

My naïve way of thinking of the axiom of choice is that it is first and foremost an axiom ensuring the existence of things. In my experience it is quite often the case that things can be defined and shown to be unique (hence “canonical”) if they exist (or non-uniqueness is controlled in some tractable way), but their existence requires additional assumptions.

It is worth mentioning that for a particular sequence of split (e.g. the one mentioned about $\mathbb{Q\to R}$) one can always generate models in which the axiom of choice holds "high enough" to ensure the splitting; but later on fails in the worst possible ways.
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Asaf KaragilaMay 30 '12 at 10:55

So my proof used the axiom of choice, but it didn't need to. If you use the fact that abelian groups are an abelian category, and use the universal properties of kernels and cokernels, you won't need AC. I had come to suspect that must be the case. Thanks.
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Joe HannonMay 30 '12 at 15:18

But wait... in your argument, you're using the fact that $f$ is the kernel of $g$, I guess? Because $f$ is mono, it must be the kernel of something, and that something must be $g$. But just to be sure, we're not going to need the axiom of choice to prove that the category of abelian groups is an abelian category, right? That every surjection of abelian groups is a cokernel is constructive as well?
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Joe HannonMay 30 '12 at 15:20

@ziggurism: I use nothing fancy and wouldn't express it this way (however, the fact that abelian groups and more generally $R$-modules form an abelian category doesn't need any choice either). Here I use that $g$ induces an isomorphism $B/\ker{g} \to C$ (it is surjective by exactness and the induced map is injective); $h$ factors over $B/\ker{g}$. This works entirely in ZF, you can just write it down: $h$ is constant on the classes $b + \ker{g}$, so it gives a well-defined map on the quotient. Similarly, $f$ is injective and gives an isomorphism from $A$ to the kernel of $g$ by exactness.
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t.b.May 30 '12 at 15:28

@ziggurism: Another side remark on this discussion is that Andreas Blass proved that the assertions "Every abelian group is projective" and "Every divisible abelian group is injective" are equivalent to the axiom of choice (each of those on its own).
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Asaf KaragilaMay 30 '12 at 21:42